What is the extraneous solution to these equations? $\dfrac{x^2 + 2}{x - 3} = \dfrac{10x - 19}{x - 3}$
Explanation: Multiply both sides by $x - 3$ $ \dfrac{x^2 + 2}{x - 3} (x - 3) = \dfrac{10x - 19}{x - 3} (x - 3)$ $ x^2 + 2 = 10x - 19$ Subtract $10x - 19$ from both sides: $ x^2 + 2 - (10x - 19) = 10x - 19 - (10x - 19)$ $ x^2 + 2 - 10x + 19 = 0$ $ x^2 + 21 - 10x = 0$ Factor the expression: $ (x - 7)(x - 3) = 0$ Therefore $x = 7$ or $x = 3$ At $x = 3$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 3$, it is an extraneous solution.